Symmetric-decreasing rearrangement of a function

Hint 1: show that if $t_1 > t_2$, then $\{ |f| > t_1\}^* \subseteq \{ |f| > t_2\}^*$.

Hint 2: use this to show that if $y\in \{ |f| > t \}^*$, then $y \in \{|f| > s\}^*$ for every $0 \leq s \leq t$. Notice that this implies that $f^*(y) \geq t$ by the definition.

Hint 3: use hint 1 again to show that if $y\not\in \{|f| > t\}^*$, $$ \sup \left\{ s \geq 0 ~~|~~ y \in \{|f| > s\}^*\right\} \leq t $$ this implies in particular $f^*(y) \leq t$ (why?).

Fix $t>0$ et $y\in \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}$. One can check that for every, $0<s< t$ one has $$\left\{x \in \mathbb{R}^n: |f(x)| > t \right\}\subset \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}$$ this entails that, \begin{equation}\label{eq-inclu t-s}\tag{I} \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}\subset \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}^{*}~~~\textrm{for all $s\in ]0,t[$}. \end{equation} this implies that,$$ \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) =1 ~~~s\in (0,t)$$

Therefore, from definition of $f^{*}$, if $y\in \{|f|>t\}^*$ then we have $$\begin{align*} f^{*}(y) &:= \int_{0}^{+ \infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds\\ &= \int_{0}^{t} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds+ \int_{t}^{+\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds\\ & = \int_{0}^{t} ds+\int_{t}^{+\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds \\ &>t. \end{align*}$$

Whence, $$\left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}\subset \left\{ x \in \mathbb{R}^n:f^{*}(x)> t \right\}.$$ On the other hand, if we suppose, $y\notin \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}$ then for all $s>0$ such that $ y\in \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}^{*}$ one has $0<s\leq t$.

Indeed, $t>s $ then from \eqref{eq-inclu t-s} $$y\in \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}$$ which is contradiction since we assumed that the converse is true. this means that,

$$\sup\left\{s>0 : y\in \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}^{*}\right\}\leq t. $$ We then deduce that, $$\begin{align*} f^{*}(y) &:= \int_{0}^{+ \infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds\\ &= \int_{0}^{t} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds+ \underbrace{\int_{t}^{+\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds}_{=0}\leq=t \end{align*}$$ that is $f^*(y)\leq t$ or that $y\notin \left\{x \in \mathbb{R}^n: f^*(x) > t \right\}$. We've just prove that,

\begin{equation}\label{eq}\tag{II} \Bbb R^n\setminus \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}\subset \Bbb R^n\setminus\left\{x \in \mathbb{R}^n: f^*(x) > s \right\}~~~\textrm{for all $s\in ]0,t[$}. \end{equation}

Which end the prove by taking the complementary.